To determine how many grams of iron can be obtained from 3.0 kg of Fe₃O₄ (magnetite), we first need to know the molar masses of the components involved.
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Calculate the molar mass of Fe₃O₄:
- Iron (Fe): approximately 55.85 g/mol
- Oxygen (O): approximately 16.00 g/mol
The molar mass of Fe₃O₄ is calculated as follows:
\[ \text{Molar mass of Fe₃O₄} = (3 \times 55.85) + (4 \times 16.00) \] \[ = 167.55 + 64.00 = 231.55 \text{ g/mol} \]
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Convert 3.0 kg of Fe₃O₄ to grams: \[ 3.0 \text{ kg} = 3000 \text{ g} \]
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Calculate the number of moles of Fe₃O₄ in 3000 g: \[ \text{Moles of Fe₃O₄} = \frac{3000 \text{ g}}{231.55 \text{ g/mol}} \approx 12.95 \text{ moles} \]
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Determine the amount of iron produced from Fe₃O₄: The chemical formula Fe₃O₄ indicates that there are 3 moles of iron (Fe) for every mole of Fe₃O₄. Thus, for 12.95 moles of Fe₃O₄, the moles of iron produced would be: \[ \text{Moles of Fe} = 12.95 \text{ moles Fe₃O₄} \times 3 \text{ moles Fe/mole Fe₃O₄} = 38.85 \text{ moles Fe} \]
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Convert moles of iron to grams: \[ \text{Mass of Fe} = \text{Moles of Fe} \times \text{Molar mass of Fe} \] \[ = 38.85 \text{ moles} \times 55.85 \text{ g/mol} \approx 2167.84 \text{ g} \]
Thus, from 3.0 kg of Fe₃O₄, approximately 2167.84 grams of iron can be obtained.